Problem: $f(x) = \dfrac{ \sqrt{ x - 3 } }{ x^2 + 9 x + 8 }$ What is the domain of the real-valued function $f(x)$ ?
Explanation: $f(x) = \dfrac{ \sqrt{ x - 3 } }{ x^2 + 9 x + 8 } = \dfrac{ \sqrt{ x - 3 } }{ ( x + 8 )( x + 1 ) }$ First, we need to consider that $f(x)$ is undefined anywhere where the radical is undefined, so the radicand (the expression under the radical) cannot be less than zero. So $x - 3 \geq 0$ , which means $x \geq 3$ Next, we also need to consider that $f(x)$ is undefined anywhere where the denominator is zero. So $x \neq -8$ and $x \neq -1$ However, these last two restrictions are irrelevant since $3 > -8$ and $3 > -1$ and so $x \geq 3$ will ensure that $x \neq -8$ and $x \neq -1$ Combining these restrictions, then, leaves us with simply $x \geq 3$ Expressing this mathematically, the domain is $\{ \, x \in \RR \mid x \geq3\, \}$.